Question

Temperature: 96.8 ^oC                         Concentration of HCl solution: 0.04998 M

	Trial 1	Trial 2	Trial 3
Initial volume of HCl (mL)	0.00	0.00	0.00
Final volume of HCl (mL)	3.15	3.84	3.37
Volume of HCl used (mL)
*Moles of Ca(OH)2 (mol)
Volume of Ca(OH)2 titrated (mL)	10.00	10.00	10.00
*Molar solubility of Ca(OH)2 (M)
*[Ca2+] (M)
*[OH-] (M)
*Ksp
*Experimental average Ksp

I have to find the answers to those with * next to them and unsure how to do those calculations.

Answer 1

Volume of HCl used = Final volume of HCl - Initial volume of HCl

= 3.15-0 = 3.15 mL

Concetration of HCl solution = 0.04998M (given)

milimoles of HCl = MHCl * VHCl

= 0.04998* 3.15 = 0.157437 milimole

Balanced chemical eqution is

Ca(OH)2 + 2HCl CaCl2 + 2H2O

Since, to neutralise 1 mole fo calcium hydroxide 2 mole of HCl is required.

Hence 1 milimoles of HCl will neutralise = 1/2 milimoles of Ca(OH)2

Hence 0.157437 milimole of ......... = 0.157437 * 1/2 milimole

= 0.0787185 ( answer of moles of Ca(OH)2

Molar solubility of Ca(OH)2 = Moles of Ca(OH)2 / Volume of Ca(OH)2 tirated

= 0.0787185* 10-3/ 10 * 10-3

= 0.00787185 M

Ca(OH)2   Ca2+ + 2 OH-

  0.00787185 M   0.00787185 M 2*0.00787185 M= 0.0157437M( [OH-])

Ksp = [Ca2+] * [OH-]2

=  0.00787185 * ( 0.0157437)2

= 1.9511 * 10-6

Experimental average Ksp of Ca(OH)2 = 6.5 * 10-6